Download Potential energy of a magnetic dipole in a magnetic field. When a

Potential energy of a magnetic dipole in a magnetic field.
When a torque is applied on an object that is free to
rotate, work is done. The incremental work done by the
field when a magnetic dipole is rotated through an angle
dθ is:
!
B
!
µ
θ
dθ
!
µ
!
B
" "
µB
U
"
µ
dW = −τ.dθ
= −µBsin θ.dθ,
where θ is the
!
angle between µ
!
and B. The work
done is equal to the decrease in potential energy of the
180!
dU = −dW = µBsin θ.dθ.
∴U = ∫ µBsin θ.dθ = −µB cosθ + U" ,
where U" is an integration constant. Choosing U = 0
when θ = 90" , then U" = 0.
! !
∴U = −µBcosθ = − µ • B ,
the potential energy of a magnetic dipole at angle θ to
the direction of a magnetic field.
360!
θ"
B
" "
−µ B
0
•
When θ = 0, U has its minimum value
(stable equilibrium).
•
system, i.e.,
θ
When θ = 180! , U has its maximum value
(unstable equilibrium).
"
Note that the torque acts to align the dipole with µ parallel
"
to B.
If the current is counter-clockwise when viewed from a
Question 26.11: A square coil with sides equal to 25.0 cm
carries a current of 2.00 A. It lies in the z = 0 plane in a
!
magnetic field B = 0.40 ˆi + 0.30 kˆ T with the current
(
)
counter-clockwise when viewed from a point on the
positive z −axis. If the coil has 6 turns what is
(a) the torque acting on the coil, and
(b) the potential energy of the coil/field system?
point on the positive z −axis, then the magnetic moment
of the coil is in the positive z direction (i.e., along kˆ ).
!
∴ µ = nIAkˆ = 6 × (2.00 A) × (0.25 m)2 kˆ
= (0.75 A ⋅ m2 )kˆ .
(a) The torque is
! ! !
τ = µ × B = (0.75 A ⋅ m2 ) kˆ × (0.40 ˆi + 0.30kˆ )T
= (0.30 N ⋅ m)ˆj.
(b) The potential energy is
! !
U = − µ • B = −(0.75 A ⋅ m2 )kˆ • (0.40 ˆi + 0.30 kˆ )T
= −0.225 J.